# A detail about algebraic topology in the proof of sheaf cohomology = singular cohomology

Recently I am reading the proof of the de Rham theorem. One step of the proof is to prove the singular cohomology $$H^i(X,\mathbb{Z})$$ is isomorphic to the sheaf cohomology $$H^i(X,\mathbb{Z})$$ of the constant sheaf $$\mathbb{Z}$$ on $$X$$. Here is a proof I found:proposition2.1

In step 3, the author gave an exact sequence as follows(here $$\mathcal{V}$$ is an open covering of $$X$$) : $$0 \rightarrow ker \pi_v \rightarrow C_{sing}^*(X,\mathbb{Z}) \xrightarrow{\pi_v} C_{sing}^{\mathcal{V},*}(X,\mathbb{Z}) \rightarrow 0$$. My question is: why is it surjective at the end of the sequence?

I know that there is a natural inclusion: $$C_*^{\mathcal{V}}(X,\mathbb{Z}) \rightarrow C_*(X,\mathbb{Z})$$, and by applying the hom functor $$Hom(\bullet,\mathbb{Z})$$ we can get a morphism $$\pi_{\mathcal{V}}: C_{sing}^*(X,\mathbb{Z}) \rightarrow C_{sing}^{\mathcal{V},*}(X,\mathbb{Z})$$, but I do not understand why it is surjective. I think it is just a simple question in algebraic topology, but I just started learning algebraic topology recently and may miss some easy facts. Thanks for any hints in advance.

Let $$i:C_* \rightarrow D_*$$ be an inclusion of chain complexes where all groups are freely generated abelian groups by sets $$SC_n$$ and $$SD_n$$ and $$i(SC_n) \subset SD_n$$ as subsets of $$C_n$$ and $$D_n$$.
Then for a map $$f:C_n \rightarrow A$$ (this is an n-cochain). Define a map $$f':D_n \rightarrow A$$ where $$f'$$ is zero on elements of $$SD_n$$ that are not in $$SC_n$$ but equal to $$f$$ on elements of $$SC_n$$.
We can do this because of the category theoretical properties of the functor which takes a set to the group freely generated by it. This property says that any homomorphism of groups $$g:FX \rightarrow A$$ where $$FX$$ is the freely generated abelian group by the set $$X$$ is equivalent to defining a function $$g:X \rightarrow A$$ on the underlying sets.
Anyways, our construction of $$f'$$ make the surjectivity of $$i^*:D^n \rightarrow C^n$$ clear since $$f' \circ i = f$$ by definition.